Inscribing Regular 15-gon in Circle

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In a given circle, it is possible to inscribe a regular 15-gon.

In the words of Euclid:

In a given circle to inscribe a fifteen-angled figure which is both equilateral and equiangular.

(The Elements: Book $\text{IV}$: Proposition $16$)


In the words of Euclid:

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.

And further, by proofs similar to those in the case of the regular pentagon, we can both inscribe a circle in the given fifteen-angled figure and circumscribe one about it.

(The Elements: Book $\text{IV}$: Proposition $16$ : Corollary)



Let $ABCD$ be the given circle.

In $ABCD$ we inscribe an equilateral triangle, one of whose vertices is at $A$.

In $ABCD$ we also inscribe a regular pentagon, one of whose vertices is at $A$.

Thus we have that $AC$ is one of the sides of the equilateral triangle, and $AB$ is one of the sides of the regular pentagon.

Let $BC$ be bisected by a line which passes through the center of the circle.

Let this line intersect the circumference of the circle at $E$.

We fit as many copies of the straight line $BE$ around the circumference of $ABCD$, starting each one at the point the previous one ends.

The resulting polygon is the required regular 15-gon.


Consider the circumference of circle $ABCD$ as divided into $15$ equal arcs.

Of these, there will be $5$ in the shorter arc $AC$, and $3$ in the shorter arc $AB$.

So there are $2$ in the shorter arc $BC$.

Once this has been bisected by the construction which produces $E$, we see that each of these parts is a copy of these $15$ equal arcs.

Then $BE$ and $EC$ are $\dfrac 1 {15}$ of the length of the circumference of $ABCD$.

Hence the result.


Historical Note

This proof is Proposition $16$ of Book $\text{IV}$ of Euclid's The Elements.